MHB [ASK] Tangent of a Circle (Again)

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The discussion revolves around finding the value of m for which the line x + my = 1 is a tangent to the circle defined by the equation x² + y² - 4x + 6y + 8 = 0. The center of the circle is identified as (2, -3) with a radius of √5. A participant suggests substituting x = 1 - my into the circle's equation and then setting the discriminant of the resulting quadratic in y to zero to find m. After performing the calculations, it is concluded that the correct value of m is 2, aligning with option D. The solution process highlights the importance of careful algebraic manipulation and verification.
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If the line x + my = 1 is a tangent of the circle $$x^2+y^2-4x+6y+8=0$$, the value of m is ...
A. -2
B. $$\frac{1}{4}$$
C. $$\frac{1}{4}$$
D. 3
E. 4

Looking at the circle's equation, the center is (2, -3) and the radius is $$\sqrt5$$. If I know the coordinate where the line meet the circle I think I can solve this by myself, but I don't know how. Should I substitute x = 1 - my to the circle equation, or is there a simpler way?
 
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Monoxdifly said:
...Should I substitute x = 1 - my to the circle equation, or is there a simpler way?

I think that's what I would do...and then equate the discriminant of the resulting quadratic in $y$ to zero. :)
 
Fine... Let's try:
$$x^2+y^2-4x+6y+8=0$$
$$(1-my)^2+y^2-4(1-my)+6y+8=0$$
$$1-2my+m^2y^2+y^2-4+4my+6y+8=0$$
$$(m^2+1)y^2+(6+2m)y+5=0$$

D = 0
$$b^2-4ac=0$$
$$(6+2m)^2-4(m^2+1)(5)=0$$
$$36+24m+4m^2-20m^2-20=0$$
$$-16m^2+24m+16=0$$
$$2m^2-3m-2=0$$
$$2m^2+m-4m-2=0$$
m(2m + 1) - 2(2m + 1) = 0
(m - 2)(2m + 1) = 0
m - 2 = 0 or 2m + 1 = 0
m = 2 or 2m = -1
After I rechecked the question, I did some typos and the option D was actually 2, so that is the answer.
 
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